If $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{i}}+2…

If $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$, $\mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$, $\mathbf{n}$ is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ and $\theta$ is the angle between $\mathbf{c}$ and $\mathbf{n}$ then $\sin \theta=$
  1. $\sqrt{\frac{2}{3}}$
  2. $\frac{\sqrt{2}}{3 \sqrt{3}}$
  3. $\frac{2}{\sqrt{3}}$
  4. $\frac{\sqrt{3}}{2}$

Solution

We have, $\mathbf{n} \perp \mathbf{a}$ and $\mathbf{n} \perp \mathbf{b}$ $ \begin{aligned} & \therefore \quad \mathbf{n}=\mathbf{a} \times \mathbf{b} \\ & =\left|\begin{array}{ccc} i & j & k \\ 1 & 2 & 3 \\ -1 & 2 & 1 \end{array}\right|=-4 i-4 j+4 k \\ & \text { Again, } \quad \sin \theta=\frac{|\mathbf{n} \times \mathbf{c}|}{|\mathbf{n}||\mathbf{c}|} \\ & \text { Now, } \quad \mathbf{n} \times \mathbf{c}=\left|\begin{array}{ccc} i & j & k \\ -4 & -4 & 4 \\ 1 & 2 & -2 \end{array}\right| \\ & =-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ & \therefore \sin \theta=\frac{|-4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}|}{|-4 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}||\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}|} \\ & =\frac{\sqrt{(-4)^2+(-4)^2}}{\sqrt{(-4)^2+(-4)^2+(4)^2} \sqrt{(1)^2+(2)^2+(-2)^2}} \\ & =\frac{\sqrt{16+16}}{\sqrt{16+16+16} \sqrt{1+4+4}}=\frac{4 \sqrt{2}}{(4 \sqrt{3}) \times 3}=\frac{\sqrt{2}}{3 \sqrt{3}} \\ & \end{aligned} $

Asked in: AP EAMCET 2018 (22 Apr Shift 1)

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